How to find the UV vertex count? is there a script to do so?
This is a little rushed, and I still need to test it a little bit more, but it seems to return the correct number of UV vertices for simple meshes. UV unwrap your mesh and run this script in object mode:

import bpy
from collections import defaultdict

mesh=bpy.context.active_object.data

uv_layer = mesh.uv_layers[0].data
verts=defaultdict(list)
loc=defaultdict(list)
count=0

for p in mesh.polygons:
    for l in p.loop_indices:
        index = mesh.loops[l].vertex_index
        uv = uv_layer[l].uv 
        co = mesh.vertices[index].co
        if (len(verts[index]) == 0):

            verts[index].append(round(uv.x,2))
            verts[index].append(round(uv.y,2))

            loc[index].append(round(co.x,2))
            loc[index].append(round(co.y,2))
            loc[index].append(round(co.z,2))

            count+=1
        else:
            uvs=verts[index]
            if(uv.x!=uvs[0] or uv.y!=uvs[1]):
                nIndex=len(verts.keys())

                while(len(verts[nIndex])&gt;0):
                    nIndex+=1

                verts[nIndex].append(round(uv.x,2))
                verts[nIndex].append(round(uv.y,2))

                loc[nIndex].append(round(co.x,2))
                loc[nIndex].append(round(co.y,2))
                loc[nIndex].append(round(co.z,2))
                count+=1   

uvco=[]

for f, b in zip(verts.items(), loc.items()):
    uvs1 = f[1]
    locs1 = b[1]

    for j, k in zip(verts.items(), loc.items()):
        uvs2 = j[1]
        locs2 = k[1]

        if (uvs1[0] == uvs2[0] and uvs1[1] == uvs2[1] and locs1[0] == locs2[0] and locs1[1] == locs2[1] and locs1[2] == locs2[2]):
            uvco.append((uvs1[0],uvs1[1],locs1[0],locs1[1],locs1[2]))


duplicatesStripped = set(tuple(i) for i in uvco) 

print("Unique UV/Vertex coordinates: %s\n" %  duplicatesStripped)   

print("There are %d UV vertices" % len(duplicatesStripped)) 


It's a little bit very messy, but I have to go to lunch now, I'll continue testing it when I get back